I ended up losing my book nd need some help Solve the following initial value ap

Question

I ended up losing my book nd need some help

Solve the following initial value applications using either separation of variables or integrating factor. A common inhabitant of human intestines is the bacterium Escherichia coll. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. Suppose that the initial population of a culture is 60 cells. Find a function for the number of cells at time t. Find the number of cells after 8 hours. When will the population reach 20,000 cells? The half-life of cesium-137 is 30 years. Suppose we have a 100-mg sample. Find a function for the amount that remains for t years. How much of the sample remains after 100 years? After how long will only 1 mg remain?

Explanation / Answer

So we start with 60 and it doubles every 20 minutes
so 60*2^(t)
t is 20 minutes. We probably want it in hours so it's 60*2^(t*3)
b) 60*2^(8*3)=60*2^24 = 1006632960
c) 60*2^(3t)=20000
20000/60=2^(3t)

log(20000/60)=log(2)*(3t)

log(20000/60)/log(2)=(3t)
log(20000/60)/log(2)/3=t
2.793 ish = t

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